A particle moving in the $xy$ -plane has velocity vector given by $v(t)=\left(\dfrac{1}{t+7},t^4\right)$ for time $t\geq 0$. What is the magnitude of the displacement of the particle between time $t=1$ and $t=3$ ? Round to the nearest tenth.
Answer: To find the magnitude of the displacement of the particle, we should first find the particle's horizontal displacement $\Delta x$ and the particle's vertical displacement $\Delta y$. Then we can find the magnitude of the displacement using the distance formula: $\text{Magnitude of displacement }=\sqrt{(\Delta x)^2+(\Delta y)^2}$ The particle's horizontal displacement can be found by taking the definite integral of the horizontal component of $v(t)$ between time $t=1$ and $t=3$ : $\Delta x=\int_{1}^{3} \dfrac{1}{t+7}\,dt=\ln(1.25)$ The particle's vertical displacement can be found by taking the definite integral of the vertical component of $v(t)$ between time $t=1$ and $t=3$ : $\Delta y=\int_{1}^{3} t^4\,dt=\dfrac{242}{5}$ Now we can find the magnitude of the displacement: $\begin{aligned} &\phantom{=}\sqrt{(\Delta x)^2+(\Delta y)^2} \\\\ &=\sqrt{\left(\ln(1.25)\right)^2+\left(\dfrac{242}{5}\right)^2} \\\\ &\approx 48.4 \end{aligned}$ In conclusion, the magnitude of the displacement of the particle between time $t=1$ and $t=3$ is $48.4$ units.